Chapter 4: SQL Basic Structure Set Operations Aggregate Functions Null Values Nested Subqueries Derived Relations Views Modification of the Database Data Definition Language 4.1 Schema Used in Examples 4.2
Basic Structure SQL is based on set and relational operations with certain modifications and enhancements A typical SQL query has the form: select A 1, A 2,..., A n from r 1, r 2,..., r m where P A i s represent attributes r i s represent relations P is a predicate. This query is equivalent to the relational algebra expression. A1, A2,..., An (σ P (r 1 x r 2 x... x r m )) The result of an SQL query is a relation. 4.3 The select Clause The select clause corresponds to the projection operation of the relational algebra. It is used to list the attributes desired in the result of a query. Find the names of all branches in the loan relation select branch-name from loan In the pure relational algebra syntax, the query would be: branch-name (loan) An asterisk in the select clause denotes all attributes select * from loan NOTE: SQL does not permit the - character in names, so you would use, for example, branch_name instead of branch-name in a real implementation. We use - since it looks nicer! NOTE: SQL names are case insensitive, meaning you can use upper case or lower case. You may wish to use upper case in places where we use bold font. 4.4
The select Clause (Cont.) SQL allows duplicates in relations as well as in query results. To force the elimination of duplicates, insert the keyword distinct after select. Find the names of all branches in the loan relations, and remove duplicates select distinct branch-name from loan The keyword all specifies that duplicates not be removed. select all branch-name from loan 4.5 The select Clause (Cont.) The select clause can contain arithmetic expressions involving the operation, +,,, and /, and operating on constants or attributes of tuples. The query: select loan-number, branch-name, amount 100 from loan would return a relation which is the same as the loan relations, except that the attribute amount is multiplied by 100. 4.6
The where Clause The where clause corresponds to the selection predicate of the relational algebra. If consists of a predicate involving attributes of the relations that appear in the from clause. The find all loan number for loans made a the Perryridge branch with loan amounts greater than $1200. select loan-number from loan where branch-name = Perryridge and amount > 1200 Comparison results can be combined using the logical connectives and, or, and not. Comparisons can be applied to results of arithmetic expressions. 4.7 The where Clause (Cont.) SQL Includes a between comparison operator in order to simplify where clauses that specify that a value be less than or equal to some value and greater than or equal to some other value. Find the loan number of those loans with loan amounts between $90,000 and $100,000 (that is, $90,000 and $100,000) select loan-number from loan where amount between 90000 and 100000 4.8
The from Clause The from clause corresponds to the Cartesian product operation of the relational algebra. It lists the relations to be scanned in the evaluation of the expression. Find the Cartesian product borrower x loan select from borrower, loan Find the name, loan number and loan amount of all customers having a loan at the Perryridge branch. select customer-name, borrower.loan-number, amount from borrower, loan where borrower.loan-number = loan.loan-number and branch-name = Perryridge 4.9 The Rename Operation The SQL allows renaming relations and attributes using the as clause: old-name asnew-name Find the name, loan number and loan amount of all customers; rename the column name loan-number as loan-id. select customer-name, borrower.loan-number as loan-id, amount from borrower, loan where borrower.loan-number = loan.loan-number 4.10
Tuple Variables Tuple variables are defined in the from clause via the use of the as clause. Find the customer names and their loan numbers for all customers having a loan at some branch. select customer-name, T.loan-number, S.amount from borrower as T, loan as S where T.loan-number = S.loan-number Find the names of all branches that have greater assets than some branch located in Brooklyn. select distinct T.branch-name from branch as T, branch as S where T.assets > S.assets and S.branch-city = Brooklyn 4.11 String Operations SQL includes a string-matching operator for comparisons on character strings. Patterns are described using two special characters: percent (%). The % character matches any substring. underscore (_). The _ character matches any character. Find the names of all customers whose street includes the substring Main. Match the name Main% select customer-name from customer where customer-street like %Main% like Main\% escape \ SQL supports a variety of string operations such as concatenation (using ) converting from upper to lower case (and vice versa) finding string length, extracting substrings, etc. 4.12
Ordering the Display of Tuples List in alphabetic order the names of all customers having a loan in Perryridge branch select distinct customer-name from borrower, loan where borrower loan-number - loan.loan-number and branch-name = Perryridge order by customer-name We may specify desc for descending order or asc for ascending order, for each attribute; ascending order is the default. E.g. order by customer-name desc 4.13 Duplicates In relations with duplicates, SQL can define how many copies of tuples appear in the result. Multiset versions of some of the relational algebra operators given multiset relations r 1 and r 2 : 1. If there are c 1 copies of tuple t 1 in r 1, and t 1 satisfies selections σ θ,, then there are c 1 copies of t 1 in σ θ (r 1 ). 2. For each copy of tuple t 1 in r 1, there is a copy of tuple Π A (t 1 ) in Π A (r 1 ) where Π A (t 1 ) denotes the projection of the single tuple t 1. 3. If there are c 1 copies of tuple t 1 in r 1 and c 2 copies of tuple t 2 in r 2, there are c 1 x c 2 copies of the tuple t 1. t 2 in r 1 x r 2 4.14
Duplicates (Cont.) Example: Suppose multiset relations r 1 (A, B) and r 2 (C) are as follows: r 1 = {(1, a) (2,a)} r 2 = {(2), (3), (3)} Then Π B (r 1 ) would be {(a), (a)}, while Π B (r 1 ) x r 2 would be {(a,2), (a,2), (a,3), (a,3), (a,3), (a,3)} SQL duplicate semantics: select A 1,, A 2,..., A n from r 1, r 2,..., r m where P is equivalent to the multiset version of the expression: Π A1,, A2,..., An (σ P (r 1 x r 2 x... x r m )) 4.15 Set Operations The set operations union, intersect, and except operate on relations and correspond to the relational algebra operations,,. Each of the above operations automatically eliminates duplicates; to retain all duplicates use the corresponding multiset versions union all, intersect all and except all. Suppose a tuple occurs m times in r and n times in s, then, it occurs: m + n times in r union all s min(m,n) times in r intersect all s max(0, m n) times in r except all s 4.16
Set Operations Find all customers who have a loan, an account, or both: (select customer-name from depositor) union (select customer-name from borrower) Find all customers who have both a loan and an account. (select customer-name from depositor) intersect (select customer-name from borrower) Find all customers who have an account but no loan. (select customer-name from depositor) except (select customer-name from borrower) 4.17 Aggregate Functions These functions operate on the multiset of values of a column of a relation, and return a value avg: average value min: minimum value max: maximum value sum: sum of values count: number of values 4.18
Aggregate Functions (Cont.) Find the average account balance at the Perryridge branch. select avg (balance) from account where branch-name = Perryridge Find the number of tuples in the customer relation. select count (*) from customer Find the number of depositors in the bank. select count (distinct customer-name) from depositor 4.19 Aggregate Functions Group By Find the number of depositors for each branch. select branch-name, count (distinct customer-name) from depositor, account where depositor.account-number = account.account-number group by branch-name Note: Attributes in select clause outside of aggregate functions must appear in group by list 4.20
Aggregate Functions Having Clause Find the names of all branches where the average account balance is more than $1,200. select branch-name, avg (balance) from account group by branch-name having avg (balance) > 1200 Note: predicates in the having clause are applied after the formation of groups whereas predicates in the where clause are applied before forming groups 4.21 Null Values It is possible for tuples to have a null value, denoted by null, for some of their attributes null signifies an unknown value or that a value does not exist. The predicate is null can be used to check for null values. E.g. Find all loan number which appear in the loan relation with null values for amount. select loan-number from loan where amount is null The result of any arithmetic expression involving null is null E.g. 5 + null returns null However, aggregate functions simply ignore nulls more on this shortly 4.22
Null Values and Three Valued Logic Any comparison with null returns unknown E.g. 5 < null or null <> null or null = null Three-valued logic using the truth value unknown: OR: (unknown or true) = true, (unknown or false) = unknown (unknown or unknown) = unknown AND: (true and unknown) = unknown, (false and unknown) = false, (unknown and unknown) = unknown NOT: (not unknown) = unknown P is unknown evaluates to true if predicate P evaluates to unknown Result of where clause predicate is treated as false if it evaluates to unknown 4.23 Null Values and Aggregates Total all loan amounts select sum (amount) from loan Above statement ignores null amounts result is null if there is no non-null amount, that is the All aggregate operations except count(*) ignore tuples with null values on the aggregated attributes. 4.24
Nested Subqueries SQL provides a mechanism for the nesting of subqueries. A subquery is a select-from-where expression that is nested within another query. A common use of subqueries is to perform tests for set membership, set comparisons, and set cardinality. 4.25 Example Query Find all customers who have both an account and a loan at the bank. select distinct customer-name from borrower where customer-name in (select customer-name from depositor) Find all customers who have a loan at the bank but do not have an account at the bank select distinct customer-name from borrower where customer-name not in (select customer-name from depositor) 4.26
Example Query Find all customers who have both an account and a loan at the Perryridge branch select distinct customer-name from borrower, loan where borrower.loan-number = loan.loan-number and branch-name = Perryridge and (branch-name, customer-name) in (select branch-name, customer-name from depositor, account where depositor.account-number = account.account-number) Note: Above query can be written in a much simpler manner. The formulation above is simply to illustrate SQL features. (Schema used in this example) 4.27 Set Comparison Find all branches that have greater assets than some branch located in Brooklyn. select distinct T.branch-name from branch ast, branch as S where T.assets > S.assets and S.branch-city = Brooklyn Same query using > some clause select branch-name from branch where assets > some (select assets from branch where branch-city = Brooklyn ) 4.28
Definition of Some Clause F <comp> some r t r s.t. (F <comp> t) Where <comp> can be: <,, >, =, 0 5 6 (5< some ) = true (read: 5 < some tuple in the relation) (5< some (5 = some 0 5 0 5 ) = false ) = true 0 (5 some 5 ) = true (since 0 5) (= some) in However, ( some) not in 4.29 Definition of all Clause F <comp> all r t r (F <comp> t) 0 5 6 (5< all ) = false (5< all (5 = all 6 10 4 5 ) = true ) = false 4 (5 all 6 ) = true (since 5 4 and 5 6) ( all) not in However, (= all) in 4.30
Example Query Find the names of all branches that have greater assets than all branches located in Brooklyn. select branch-name from branch where assets > all (select assets from branch where branch-city = Brooklyn ) 4.31 Test for Empty Relations The exists construct returns the value true if the argument subquery is nonempty. exists r r Ø not exists r r = Ø 4.32
Example Query Find all customers who have an account at all branches located in Brooklyn. select distinct S.customer-name from depositor as S where not exists ( (select branch-name from branch where branch-city = Brooklyn ) except (select R.branch-name from depositor as T, account as R where T.account-number = R.account-number and S.customer-name = T.customer-name)) (Schema used in this example) Note that X Y = Ø X Y Note: Cannot write this query using = all and its variants 4.33 Test for Absence of Duplicate Tuples The unique construct tests whether a subquery has any duplicate tuples in its result. Find all customers who have at most one account at the Perryridge branch. select T.customer-name from depositor as T where unique ( select R.customer-name from account, depositor as R where T.customer-name = R.customer-name and R.account-number = account.account-number and account.branch-name = Perryridge ) (Schema used in this example) 4.34
Example Query Find all customers who have at least two accounts at the Perryridge branch. select distinct T.customer-name from depositor T where not unique ( select R.customer-name from account, depositor as R where T.customer-name = R.customer-name and R.account-number = account.account-number and account.branch-name = Perryridge ) (Schema used in this example) 4.35 Views Provide a mechanism to hide certain data from the view of certain users. To create a view we use the command: create view v as <query expression> where: <query expression> is any legal expression The view name is represented by v 4.36
Example Queries A view consisting of branches and their customers create view all-customer as (select branch-name, customer-name from depositor, account where depositor.account-number = account.account-number) union (select branch-name, customer-name from borrower, loan where borrower.loan-number = loan.loan-number) Find all customers of the Perryridge branch select customer-name from all-customer where branch-name = Perryridge 4.37 Derived Relations Find the average account balance of those branches where the average account balance is greater than $1200. select branch-name, avg-balance from (select branch-name, avg (balance) from account group by branch-name) as result (branch-name, avg-balance) where avg-balance > 1200 Note that we do not need to use the having clause, since we compute the temporary (view) relation result in the from clause, and the attributes of result can be used directly in the where clause. 4.38
With Clause With clause allows views to be defined locally to a query, rather than globally. Analogous to procedures in a programming language. Find all accounts with the maximum balance with max-balance(value) as select max (balance) from account select account-number from account, max-balance where account.balance = max-balance.value 4.39 Complex Query using With Clause Find all branches where the total account deposit is greater than the average of the total account deposits at all branches with branch-total (branch-name, value) as select branch-name, sum (balance) from account group by branch-name with branch-total-avg(value) as select avg (value) from branch-total select branch-name from branch-total, branch-total-avg where branch-total.value >= branch-total-avg.value 4.40
Modification of the Database Deletion Delete all account records at the Perryridge branch delete from account where branch-name = Perryridge Delete all accounts at every branch located in Needham city. delete from account where branch-name in (select branch-name from branch where branch-city = Needham ) delete from depositor where account-number in (select account-number from branch, account where branch-city = Needham and branch.branch-name = account.branch-name) (Schema used in this example) 4.41 Example Query Delete the record of all accounts with balances below the average at the bank. delete from account where balance < (select avg (balance) from account) Problem: as we delete tuples from deposit, the average balance changes Solution used in SQL: 1. First, compute avg balance and find all tuples to delete 2. Next, delete all tuples found above (without recomputing avg or retesting the tuples) 4.42
Modification of the Database Insertion Add a new tuple to account insert into account values ( A-9732, Perryridge,1200) or equivalently insert into account (branch-name, balance, account-number) values ( Perryridge, 1200, A-9732 ) Add a new tuple to account with balance set to null insert into account values ( A-777, Perryridge, null) 4.43 Modification of the Database Insertion Provide as a gift for all loan customers of the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account insert into account select loan-number, branch-name, 200 from loan where branch-name = Perryridge insert into depositor select customer-name, loan-number from loan, borrower where branch-name = Perryridge and loan.account-number = borrower.account -number The select from where statement is fully evaluated before any of its results are inserted into the relation (otherwise queries like insert into table1 select * from table1 would cause problems 4.44
Modification of the Database Updates Increase all accounts with balances over $10,000 by 6%, all other accounts receive 5%. Write two update statements: update account set balance = balance 1.06 where balance > 10000 The order is important update account set balance = balance 1.05 where balance 10000 Can be done better using the case statement (next slide) 4.45 Case Statement for Conditional Updates Same query as before: Increase all accounts with balances over $10,000 by 6%, all other accounts receive 5%. update account set balance = case when balance <= 10000 then balance *1.05 else balance * 1.06 end 4.46
Update of a View Create a view of all loan data in loan relation, hiding the amount attribute create view branch-loan as select branch-name, loan-number from loan Add a new tuple to branch-loan insert into branch-loan values ( Perryridge, L-307 ) This insertion must be represented by the insertion of the tuple into the loan relation ( L-307, Perryridge, null) Updates on more complex views are difficult or impossible to translate, and hence are disallowed. Most SQL implementations allow updates only on simple views (without aggregates) defined on a single relation 4.47 Transactions A transaction is a sequence of queries and update statements executed as a single unit Transactions are started implicitly and terminated by one of commit work: makes all updates of the transaction permanent in the database rollback work: undoes all updates performed by the transaction. Motivating example Transfer of money from one account to another involves two steps: deduct from one account and credit to another If one steps succeeds and the other fails, database is in an inc onsistent state Therefore, either both steps should succeed or neither should If any step of a transaction fails, all work done by the transaction can be undone by rollback work. Rollback of incomplete transactions is done automatically, in case of system failures 4.48
Transactions (Cont.) In most database systems, each SQL statement that executes successfully is automatically committed. Each transaction would then consist of only a single statement Automatic commit can usually be turned off, allowing multistatement transactions, but how to do so depends on the database system Another option in SQL:1999: enclose statements within begin atomic end 4.49 Data Definition Language (DDL) Allows the specification of not only a set of relations but also information about each relation, including: The schema for each relation. The domain of values associated with each attribute. Integrity constraints The set of indices to be maintained for each relations. Security and authorization information for each relation. The physical storage structure of each relation on disk. 4.50
Domain Types in SQL char(n). Fixed length character string, with user-specified length n. varchar(n). Variable length character strings, with user-specified maximum length n. int. Integer (a finite subset of the integers that is machine-dependent). smallint. Small integer (a machine-dependent subset of the integer domain type). numeric(p,d). Fixed point number, with user-specified precision of p digits, with n digits to the right of decimal point. real, double precision. Floating point and double-precision floating point numbers, with machine-dependent precision. float(n). Floating point number, with user-specified precision of at least n digits. Null values are allowed in all the domain types. Declaring an attribute to be not null prohibits null values for that attribute. create domain construct in SQL-92 creates user-defined domain types create domain person-name char (20) not null 4.51 Date/Time Types in SQL (Cont.) date. Dates, containing a (4 digit) year, month and date E.g. date 2001-7-27 time. Time of day, in hours, minutes and seconds. E.g. time 09:00:30 time 09:00:30.75 timestamp: date plus time of day E.g. timestamp 2001-7-27 09:00:30.75 Interval: period of time E.g. Interval 1 day Subtracting a date/time/timestamp value from another gives an interval value Interval values can be added to date/time/timestamp values Can extract values of individual fields from date/time/timestamp E.g. extract (year from r.starttime) Can cast string types to date/time/timestamp E.g. cast <string-valued-expression> as date 4.52
Create Table Construct An SQL relation is defined using the create table command: create table r (A 1 D 1, A 2 D 2,..., A n D n, (integrity-constraint 1 ),..., (integrity-constraint k )) r is the name of the relation each A i is an attribute name in the schema of relation r D i is the data type of values in the domain of attribute A i Example: create table branch (branch-name char(15) not null, branch-city char(30), assets integer) 4.53 Integrity Constraints in Create Table not null primary key (A 1,..., A n ) check (P), where P is a predicate Example: Declare branch-name as the primary key for branch and ensure that the values of assets are nonnegative. create table branch (branch-namechar(15), branch-city char(30) assets integer, primary key (branch-name), check (assets >= 0)) primary key declaration on an attribute automatically ensures not null in SQL-92 onwards, needs to be explicitly stated in SQL-89 4.54
Drop and Alter Table Constructs The drop table command deletes all information about the dropped relation from the database. The after table command is used to add attributes to an existing relation. All tuples in the relation are assigned null as the value for the new attribute. The form of the alter table command is alter table r add A D where A is the name of the attribute to be added to relation r and D is the domain of A. The alter table command can also be used to drop attributes of a relation alter table r drop A where A is the name of an attribute of relation r Dropping of attributes not supported by many databases 4.55 End of Chapter